Question

Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5×10−2 If you start with 26.0 mL of a 0.904 M solution of NaIO4, and then dilute it with water to 500.0 mL, what is the concentration of H4IO−6 at equilibrium?

Answer 1

0.744 M

Step-by-step explanation:

IO⁻⁴(aq) + 2H₂O(l) ⇌ H₄IO⁻⁶(aq)

Kc = 3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

First let's calculate the new concentration of IO⁻⁴ at equilibrium:

0.904 M * 26.0 mL / 500.0 mL = 0.047 M = [IO⁻⁴]

Now we can calculate [H₄IO⁻⁶] using the formula for Kc:

3.5×10⁻²= [H₄IO⁻⁶] / [IO⁻⁴]

3.5×10⁻²= [H₄IO⁻⁶] / 0.047 M

[H₄IO⁻⁶] = 0.744 M